Many applications exist for step-up voltage converters capable of delivering high-currents. One such application is the need to drive LEDs at high currents for the camera flash function. The brightness of white LEDs is proportional to current. Currents may range from a few hundred milliamperes to several amperes. At high currents the voltage drop across a white LED can be more than three to four volts. Special high-intensity LEDs for high brightness camera flash applications exhibit even higher voltages, voltages higher than the voltage of a single cell lithium ion battery. In order to drive such high brightness LEDs, the battery voltage must be stepped up to a higher potential.
While charge pump converters can be used up to one half ampere, the input current of a charge pump is multiplied by the voltage conversion ratio. A doubler charge pump produces an output voltage double its input but requires an input current at least double its output. So at one-half ampere LED current, the input current of the charge pump LED driver exceeds one ampere. At higher currents, the input current demand of a charge pump becomes excessive. In such instances a boost type switching regulator is preferred.
As shown in FIG. 1A, boost converter and switching regulator 1 comprises power MOSFET 2, inductor 3, Schottky rectifier 4, output capacitor 5 and PWM controller 10. The converter drives camera flash LED 6 with the flash current and duration set by MOSFET 7, resistor 8, and flash control timing circuit 9. Flash operation comprises pre-charging capacitor 5 before driving LED 6 at high currents. Pre-charging involves turning on low-side MOSFET 2 and magnetizing inductor 3 with current conduction path (1) for some duration ton then turning off low-side MOSFET 2, whereby inductor 3 drives the potential Vx above the converter's output, i.e. above that of the potential across capacitor 5. During this interval, Schottky diode 4 becomes forward biased and transfers energy from inductor 3 to capacitor 5 as shown by current-conduction path (2).
The converter alternates between conduction states (1) and (2) until the boost output voltage Vboost reaches its target value set by PWM controller 10 reacting in response to feedback voltage VFB. This pre-charging operating sequence is illustrated by voltage ramp 21 illustrated in graph 20 of FIG. 1B, stabilizing at a potential Vready 22 at time t1. During this interval flash MOSFET 7 remains biased in an off condition, the current ILED in light emitting diode 6 is zero, and LED 6 does not illuminate.
At time t2, flash controller 9 turns on power MOSFET 7 and current flows from the boost converter's output through LED 6, conducting MOSFET 7, and current setting resistor 8 to ground along the conduction path designated by the dashed arrow (3). Assuming MOSFET 7 has an on-state resistance of RDS(on) and resistor Rset has a resistance substantially greater than the MOSFET's on-resistance, the current flowing during this interval is then given by the relation
      I    LED    =                              V          boost                -                  V                      f            ⁡                          (              LED              )                                                            R                      DS            ⁡                          (              on              )                                      +                  R          set                      ≈                            V          boost                -                  V                      f            ⁡                          (              LED              )                                                  R        set            
Assuming that Vf(LED) at high current is 4.5V, the Vboost is biased to 4.6V, and using a resistor value of Rset=0.1Ω, then ILED=1 A.
Although a 1 A current is thirty times the current normally used for lighting LEDs, high brightness flash LEDs typically comprise an array of parallel LEDs. Moreover in a camera flash application, the LED conduction duration is limited to a few hundred milliseconds. By controlling the flash time, the total energy dissipated by the LEDs is limited and the LED array is not damaged.
During the flash interval, the ILED current 24 decays in proportion to the sagging Vboost voltage 23 present on capacitor 5. During the flash interval, converter 1 operating in fixed frequency operation naturally alternates between discharge path (3) and magnetizing condition (1), with the converter attempting to minimize the voltage sag on inductor 5. The degree of voltage sag depends on the magnitude of inductance in inductor 3 and the magnitude of capacitance of capacitor 5.
Operating in variable frequency mode during the flash, converter 1 may remain in the discharging state with conduction path (3) for an extended duration. If the inductor current drops too low, however, the LED brightness will fall to unacceptable brightness levels. To avoid this problem even in a variable frequency mode, converter 1 must occasionally return to condition (1) to magnetize inductor 3 and to, at least in part, restore its current.
At time t3, when the flash interval is complete, MOSFET 7 is turned off and ILED drops to zero, during which the converter returns to alternating between states (1) and (2). With ILED=0, the value of Vboost recovers back to its ready state, as shown at time t4.
One major limitation of an inductive boost converter is its need to draw high currents from the battery. In FIG. 1A, all current paths (1), (2), and (3) flows from the battery terminal Vbatt. High currents flowing from and through the battery can cause a number of problems including sudden and unwanted voltage transients. This issue is further illustrated in flash driver schematic 30 in FIG. 2A where battery pack 31 powers boost converter 35 to drive LED 38. Power MOSFET 39 controls the flash duration while Vboost and the value of resistance 40 sets LED current and brightness. Boost converter 36 includes input filter capacitor 34, inductor 36, and output filter capacitor 37.
The input power delivered to boost converter 34 is equal to the power delivered to the load and any additional power needed to operate converter 35. Assuming as a best case 100% efficiency of boost converter 35 then its power input must equal its power output so thatPIN=POUT=IbattVbatt=VboostIboost 
If the charge stored on output capacitor 37 is negligible compared to the total current required to fire flash LED 38 then it follows Iboost≈ILED and therefore
      I    batt    =                    I        boost            ⁢                        V          boost                          V          batt                      ≈                  I        LED            ⁢                        V          boost                          V          batt                    
Since by definition for a boost converter Vboost>Vbatt, then the converter's average input must exceed the current load current ILED during the flash. For a 5V LED and a single cell lithium ion battery at 3.6V, the average input current to the converter is roughly 40% higher. So to achieve an average of 1 A an LED flash demands a 1.4 A current from the battery.
Referring again to FIG. 2A, battery pack 31 comprises electrochemical cell 32 with voltage Vcell and internal resistance 33. Resistance 33 may comprise resistance associated with a battery's internal electrodes, electrochemical resistance, and any resistance introduced by protection electronics within the battery pack itself. In a single cell lithium ion battery pack for example, Vcell may comprise a voltage 4.2V to 3V and the resistance Rbattery of resistor 33 can be as high as 500 mΩ. Ideally the battery's resistance should be zero. Because the resistance is not negligible, high currents can cause excess heating inside the battery pack and an unwanted voltage drop within the battery pack itself. Assuming a battery current Ibatt, the battery's voltage at its terminals is given byVbatt=Vcell−RbattIbatt 
If, for example a 1.4 A current flows through a 500 mΩ pack, a 700 mV volt drop will occur in the pack. If the battery's cell is partially discharged to 3.5V, a sudden current spike to 1.4 A can cause the battery's terminal voltage to drop 2.8V. Such a transient is illustrated in graph 45 of FIG. 2B, where the unloaded battery starts with its cell voltage 46 then drops to a lower voltage 47. If this voltage is too low, it may trigger under-voltage protection in circuitry being powered by the battery, shutting off the electrical load and interrupting the product's operation. As soon as the current is interrupted, the voltage jumps up 48 to its unloaded voltage. The user perceives the impact of this unwanted voltage transient as spurious or unreliable product operation, or in the very least, as a shortened battery life.
Aside from issues of voltage transients from high currents, the overall power demanded by the LED during a camera flash puts additional requirements on a boost converter's operation.
LED Drive Energy and Power Requirements
The power requirements of the boost converter driving high current LEDs as a camera flash is given by the relationP=ILED·Vf(LED)=Plight+Pheat 
The total power consumed by the LED is then 1 A times 4.5V or 4.5 W. As an energy conversion device, the optical efficiency of the LED is ηλ, thenPheat=(100%−ηλ)(ILED·Vf(LED))and assuming a 60% energy conversion efficiency Pheat=40% (4.5 W)=1.8 W of peak thermal dissipation divided among several LEDs in the array. Since the flash duty factor is very low, typically less than a few percent, the average power dissipation is only hundred milliwatts so that overheating and excessive LED temperatures are not critical.
Furthermore the thermal energy absorbed by the LED in a single-pulse is of limited duration, the LED is not damaged during a single pulse even at high currents. The absorbed thermal energy is given by the relationEheat=Pheat·tflash=(100%−ηλ)(ILED·Vf(LED))tflash 
For example if the flash time tflash is several hundred milliseconds, e.g. 200 milliseconds, the total energy dissipated as heat is E=(1.8 W)(0.2 s)=360 mJ, not a large amount of energy for a short transient pulse.
The total energy supplied by the converter to the LED during the flash is larger, however, since it includes energy converted to the light output as well as to Joule heating. GivenE=Ptotaltflash=(ILED·Vf(LED))tflash 
Then at 4.5V and 1 A, a 200 msec pulse requires a boost converter to deliver 900 mJ to power the flash operation.
In a boost converter 1 the energy stored in the output capacitor 5 is
  E  =            1      2        ⁢    C    ⁢                  ⁢          V      boost      2      
Assuming C=4.7 uF, the capacitor stores only around 50 μj. The energy stored in inductor 3 is given by
  E  =            1      2        ⁢          LI      L      2      
At 1 A, the energy stored in a 4.7 uH inductor is then only 2.3 μj. Both inductor 3 and capacitor 5 store too little energy to power the entire duration of flash. This means that using reasonable values of inductance and capacitance there is not enough energy stored in a boost converter to power the entire flash and instead the switching regulator must keep switching and transferring energy during the entire flash pulse. Operation of a switching converter however continuously draws power from the battery at an average current higher than the flash LED's current.
Super-Cap Flash Technique
To avoid the need for drawing current from the battery during a flash, a large storage capacitor can be used to supply the entire flash transient. Such a solution 50 is illustrated in FIG. 3 comprising battery pack 51, large storage capacitor 50, boost converter 56 with inductor 57 and filter capacitor 58, LED 59, power MOSFET 60 and current setting resistor 61. Battery pack 51 includes electrochemical cell 52 and internal resistance 53. It is assumed in this example that capacitor 55, herein referred to as a super capacitor or super-cap for short, has a capacitance many orders of magnitude above that of filter capacitor 58. Operation involves charging capacitor 55 then using the energy stored on capacitor 55 to power LED 59 through boost converter 56 during camera flash operation.
To estimate the magnitude of the energy that must be stored on super-cap 55, the relation
  E  =                    P        total            ⁢              t        flash              =                            (                                    I              LED                        ·                          V                              f                ⁡                                  (                  LED                  )                                                              )                ⁢                  t          flash                    =                        1          2                ⁢                  C          sc                ⁢                  V          sc          2                                    must be satisfied. Rearranging terms gives        
      C    sc    =            2              V        sc        2              ⁡          [                        (                                    I              LED                        ·                          V                              f                ⁡                                  (                  LED                  )                                                              )                ⁢                  t          flash                    ]      
Super capacitors however can only be charged to low voltages, e.g. to 4V without damaging their internal dielectrics. Assuming the flash energy requirement of 900 mJ calculated previously the resulting capacitance required is then approximately 110 mF, i.e. over one-tenth of a Farad, four orders of magnitude greater than normal capacitors. Super capacitors up to one Farad are now commercially available. Several disadvantages of super capacitors, however, are that they are expensive and large, possibly too large to be useful in space conscious digital still cameras and camera phones.
Another complication of super-caps is that they cannot be charged directly from the battery. If an uncharged super-cap is connected directly across a battery, it behaves identically to a dead short and may damage the battery. Instead the charging current must be regulated by additional circuitry 54, adding and cost and complexity to the super-cap camera flash LED solution.
So while the super-cap solves the issue of drawing excessive currents from a battery pack during an LED camera flash it is expensive, large, and complex to operate. What is needed is a means to drive an LED at high currents and at voltages higher than the battery's voltage without drawing high or excessive currents from the battery during flash operation.